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Genetic 342 A&B Kenneth G Wilson Professory of Botany |
Trihybrid cross Assuming Independent Assortment
There are two possible solutions
Classical Solution
In Drosophila the following wild type alleles are used for analysis v (vermilion eyes), cv (cross-veinless) and ct (cut or snipped wing edges). Flies doubly recessive and homozygous (v+v+,cvcv,ctct) are crossed with flies singly recessive and homozygous (vv, cv+cv+, ct+ct+). Female progeny that are vv+, cvcv+, ctct+ are then test crossed with the tripley recessive males (vv, cvcv, ctct)
P1=v+v+,cvcv,ctct x vv, cv+cv+, ct+ct+
F1 =Wild Type - V+V, cv+cv, ct+ct x vv, cvcv, ctct (gametes v, cv, ct)
F1 gamete types | Obs. | Description | ||
v |
cv+ |
ct+ |
580 |
Parental |
v+ |
cv |
ct |
592 |
Parental |
v |
cv |
ct+ |
45 |
Crossover Region I |
v+ |
cv+ |
ct |
40 |
Crossover Region I |
v |
cv |
ct |
89 |
Crossover Region II |
v+ |
cv+ |
ct+ |
94 |
Crossover Region II |
v |
cv+ |
ct |
3 |
Double Crossover |
v+ |
cv |
ct+ |
5 |
Double Crossover |
1448 |
Finding the center gene by comparing parental to double crossover.
580+59 2 |
Parental |
45+40 |
Crossover Region I |
89+94 |
Crossover Region II |
3+5 |
Double Crossover |
1448 |
C/O |
Chromosome region |
% Recombination |
45+40 |
Crossover Region I |
[(45+40)+(3+5)]x100/1448=6.4% |
89+94 |
Crossover Region II |
[(89+94)+(3+5)]x100/1448=13.2% |
3+5 |
Double Crossover |
|
Map data
v<---13.2--->ct<--6.4-->cv
Can we calculate the Interference?
The double cross overs are always the smallest in number. In this problem that is 5+3. The expected frequency of double crossovers can be calculated by multiplying the frequency of a crossover between v and ct by the frequency of a crossover between ct and cv. 0.132 x 0.064 = 0.0084
The expected number of double crossovers is equal to the expected frequence of double crossovers times the total number of individuals . 0.0084 x 1448 = 12
Interference (I) = 1 -([# of dble c/o]/[exp # of dble c/o]) = 1- (8/12) = 0.33 }
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